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Serga [27]
3 years ago
14

Please help!!! zoom in to read it clearly

Physics
1 answer:
Artist 52 [7]3 years ago
4 0

Answer:

imma try nd help cuz i havent seen anybody answer so

1.sara

2.same for both

3.tom

Explanation:

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Eld a distance r1 from P. The second particle is then released. Determine its speed when it is a distance r2 from P. Let q = 3.1
zheka24 [161]

Answer:

v_{f}=1721.1m/s

Explanation:

Given data

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As we know that:

dK=-dU\\(1/2)mv_{f}^{2}-(1/2)mv_{i}^{2}=-(\frac{kq^{2} }{r_{f}}-\frac{kq^{2} }{r_{i}} )\\    (1/2)*(47*10^{-6}kg )v_{f}^{2}-(1/2)*(47*10^{-6}kg)(0)=-[(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{2.5*10^{-3}}-(\frac{(9*10^{9}(3.1*10^{-6})^{2}  }{0.83*10^{-3}} )]\\2.35*10^{-5} v_{f}^{2}=69.61\\v_{f}=\sqrt{\frac{69.61}{2.35*10^{-5}} }\\v_{f}=1721.1m/s

5 0
3 years ago
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pantera1 [17]

Answer:

200 K

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0 °C = 273 K

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